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8/10/2019 425 Serviceability Chp 6 S11
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Chp-6:Lecture Goals
Serviceability
Deflection calculation
Deflection example
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Structural Design Profession is concerned with:
Limit States Philosophy:Strength Limit State
(safety-fracture, fatigue, overturning buckling
etc.-)Serviceability
Performance of structures under normal service
load and are concerned the uses and/oroccupancy of structures
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;
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:Time dependent Factor
:Amplification factor
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Solution:
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See Example 2.2
Ma: Max.Service
Load Moment
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Reinforced Concrete Sections - Example
Given a doubly reinforced beam with h = 24 in, b = 12 in.,
d = 2.5 in. and d = 21.5 in. with 2# 7 bars in compression
steel and 4 # 7 bars in tension steel. The materialproperties are fc= 4 ksi and fy= 60 ksi.
Determine Igt, Icr, Mcr(+), Mcr(-), and compare to the NA of
the beam.
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Reinforced Concrete Sections - Example
The components of the beam
2 2
s
2 2
s
c c
2 0.6 in 1.2 in
4 0.6 in 2.4 in
1 k57000 57000 40001000 lb
3605 ksi
A
A
E f
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Reinforced Concrete Sections - Example
The compute the n value and the centroid, I uncracked
n area (in ) n*area (in ) yi(in) yi*n*area (in2) I (in ) d (in) d *n*area(in )
A's 7.04 1.2 8.448 2.5 21.12 - -9.756 804.10
As 7.04 2.4 16.896 21.5 363.26 - 9.244 1443.75
Ac 1 288 288 12 3456.00 13824 -0.256 18.89
313.344 3840.38 13824 2266.74
s
c
29000 ksi
8.043605 ksi
E
n E
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Reinforced Concrete Sections - Example
The compute the centroid and I uncracked
3
i i i2
i i
2 4 4
i i i i
4
3840.38 in12.26 in.
313.34 in
I 13824 in 2266.7 in
16090.7 in
y n Ay
n A
I d n A
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Reinforced Concrete Sections - Example
The compute the centroid and I for a cracked doublyreinforced beam.
0
212212 ssss2
b
dnAAny
b
nAAny
2 2
2
2 2
s
2
2 7.04 1.2 in 2 8.04 2.4 in
12 in.
2 7.04 1.2 in 2 8.04 2.4 in 21.5 in.0
12 in.
4.624 72.664 0
y y
y y
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Reinforced Concrete Sections - Example
The compute the centroid for a cracked doubly reinforced
beam.
2
4.624 72.664 0y y
2
4.624 4.624 4 72.664
2
6.52 in.
y
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Reinforced Concrete Sections - Example
The compute the moment of inertia for a cracked doubly
reinforced beam.
2s2
s
3
cr 1
3
1ydnAdyAnybI
3
cr
22
22
4
112 in. 6.52 in.
3
7.04 1.2 in 6.52 in. 2.5 in.
8.04 2.4 in 21.5 in. 6.52 in.
5575.22 in
I
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Reinforced Concrete Sections - Example
The critical ratio of moment of inertia
4
cr
4g
cr g
5575.22 in
0.34616090.7 in
0.35
I
I
I I
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Reinforced Concrete Sections - Example
Find the components of the beam
c c
s
s s s
2
s s s c
0.85 0.85 4 ksi 12 in. 0.85 34.68
2.5 in. 0.00750.003 0.003
0.0075 217.529000 0.003 87
217.50.85 1.2 in 87
261100.32
C f ba c c
cc c
f Ec c
C A f f c
c
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Reinforced Concrete Sections - Example
Find the components of the beam
2
c s
2
2
2.4 in 60 ksi 144 k
261144 k 34.68 100.32 34.68 43.68 261 0
43.68 43.68 4 261 34.68
2 34.68
3.44 in.
T
T C C
c c cc
c
The neutral axis
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Reinforced Concrete Sections - Example
The strain of the steel
Note: At service loads, beams are assumed to act
elastically.
s
s
3.44 in. 2.5 in.0.003 0.0008 0.00207
3.44 in.
21.5 in. 3.44 in. 0.003 0.0158 0.002073.44 in.
3.44 in.
y 6.52 in.
c
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Reinforced Concrete Sections - Example
Using a linearly varying and s= E along the NAis the centroid of the area for an elastic center
The maximum tension stress in tension is
r c7.5 7.5 4000474.3 psi 0.4743 ksi
f f
My
I
s
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Calculate the Deflections
(1) Instantaneous (immediate) deflections
(2) Sustained load deflection
Instantaneous Deflections
due to dead loads( unfactored) , live, etc.
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Calculate the DeflectionsInstantaneous Deflections
Equations for calculating Dinstfor common cases
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Calculate the DeflectionsInstantaneous Deflections
Equations for calculating Dinstfor common cases
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Calculate the DeflectionsInstantaneous Deflections
Equations for calculating Dinstfor common cases
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Calculate the DeflectionsInstantaneous Deflections
Equations for calculating Dinstfor common cases
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Sustained Load Deflections
Creep causes an increase
in concrete strainCurvature
increases
Compression steel
present
Increase in compressive
strains cause increase in
stress in compression
reinforcement (reducescreep strain in concrete)Helps limit thiseffect.
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Sustained Load Deflections
Sustain load deflection = l Di
Instantaneous deflection
l
501 ACI 9.5.2.5
bd
As
at midspan for simple and continuous beams
at support for cantilever beams
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Sustained Load Deflections
For dead and live loads
total DL inst LL inst
DL L.T. LL L.T.
D D D
D D
DL and LL may have different factors for LT ( long
term ) Dcalculations
instDLtotal
componentsN/S
ofattachmentaftertotal DDD
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Sustained Load Deflections
The appropriate value of Icmust be used to calculate
D at each load stage.
instDLinstLL
instDL
DDD Some percentage of DL (if given)
Full DL and LL
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Serviceabil i ty Load Deflections - Example
fc= 5 ksi
fy= 60 ksi
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Serviceabil i ty Load Deflections - Example
Part I
Determine whether the floor beam meets the ACI Code maximum
permissible deflection criteria. (Note: it will be assumed that it is
acceptable to consider the effective moments of inertia at location
A and B when computing the average effective moment of inertia
for the span in this example.)
Part IICheck the ACI Code crack width provisions at midspan of the
beam.
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Serviceabil i ty Load Deflections - Example
Compute the centroid and gross moment of inertia, Ig.
b h Area yi Ai * yi
Flange 84 4.5 378 17.75 6709.5
Web 15.5 12 186 7.75 1441.5
564 8151
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Serviceabil i ty Load Deflections - Example
The moment of inertia3
2
8151 in14.45 in 14.5 in
564 in
i i
i
y Ay
A
3 2
g
3 22
3 22
4 4
1
12
184 in 4.5 in 378 in 17.75 in - 14.45 in
121
12 in 15.5 in 186 in 7.75 in - 14.45 in12
16,950 in 16900 in
I bh Ad
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Serviceabil i ty Load Deflections - Example
The moment capacity
r c
c c
4
r g
cr -
t -
4
r g
cr +
t +
7.5 7.5 5000 530 psi
E 57000 57000 5000 /1000 lbs=4030 ksi
530 psi 16900 inM 1610 k-in
y 5.55 in 1000 lbs/kip
530 psi 16900 inM 618 k-iny 14.5 in 1000 lbs/kip
f f
f
f I
f I
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Serviceabil i ty Load Deflections - Example
Determine bending moments due to initial load (0.85DL) The ACI moment coefficients will be used to
calculate the bending moments Since the loading is not
patterned in this case, This is slightly conservative
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Serviceabil i ty Load Deflections - Example
The moments at the two locations
2
A D n
2
2
B D n
2
0.85 /11
0.85*1.00k/ft* 33.67 ft /1187.6 k-ft 1050 k-in
0.85 /16
0.85*1.00k/ft* 33.67 ft /16
60.2 k-ft 723 k-in
M w l
M w l
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Serviceabil i ty Load Deflections - Example
Moment at C will be set equal to Mafor simplicity, asgiven in the problem statement.
A cr -
g
B cr +
cr
1050 k-in M
618 k-in Use I @ midspan
M
M
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Serviceabil i ty Load Deflections - Example
Assume Rectangular Section Behavior and calculate the
areas of steel and ratio of Modulus of Elasticity
s
c
2 2
s
2 2
s
29000 ksi7.2
4030 ksi3#5 3 0.31 in 0.93 in
4#7 4 0.6 in 2.40 in
En
EA
A
S f
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Serviceabil i ty Load Deflections - Example
Calculate the center of the T-beam
s s s s2
2 2
2
2 2
2
2 1 2 2 1 2 0
2 6.2 0.93 in 2 7.2 2.4 in
84 in.
2 6.2 0.93 in 2.5 in. 2 7.2 2.4 in 17.5 in.0
84 in.
0.549 7.54 0
0.549 30.472.49 in.
2
n A nA n A d nA d y yb b
y y
y y
y
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Serviceabil i ty Load Deflections - Example
The centroid is located at the As< 4.5 in. = tf Userectangular section behavior
2 23
s scr +
3
22
22
4
11
3
184 in 2.49 in
3
6.2 0.93 in 2.5 in 2.49 in
7.2 2.4 in 17.5 in 2.49 in
4330 in
I by n A y d nA d y
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Serviceabil i ty Load Deflections - Example
The moment of inertia at midspan
3 3
cr cr g cre midspan
a a
3
4
3
4
4
1
618 k-in16900 in
723 k-in
618 k-in1 4330 in723 k-in
12200 in
M MI I I
M M
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Serviceabil i ty Load Deflections - Example
Calculate average effective moment of inertia, Ie(avg)forinterior span (for 0.85 DL) For beam with two ends
continuous and use Igfor the two ends.
e1 e2e avg e mid
4
4 4
4
0.7 0.15
0.7 12200 in
0.15 16900 in 16900 in
13600 in
I I I I
S i bil i t L d D fl ti E l
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Serviceabil i ty Load Deflections - Example
Calculate instantaneous deflection due to 0.85 DL:
Use the deflection equation for a fixed-fixed beam but usethe span length from the centerline support to centerline
support to reasonably approximate the actual deflection.
4 34
DL inst 4
0.85 1.00 k/ft 35 ft 12 in/ft
384 384 4030 ksi 13600 in
0.105 inches
l
EI
D
S i bil i t L d D fl ti E l
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Serviceabil i ty Load Deflections - Example
Calculate additional short-term Deflections (full DL & LL)
22
D L n
A
22
D L n
B
1.62 k/ft 33.67 ft
11 11
167 k-ft 2000 k-in1.62 k/ft 33.67 ft
16 16
115 k-ft 1380 k-in
lM
lM
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Serviceabil i ty Load Deflections - Example
Calculate additional short-term Deflections (full DL & LL)
Let Mc= Ma= - 2000 k-in for simplicity see problem
statement
c A cr -
B cr +
2000 k-in 1610 k-in
cracking at supports
1380 k-in 618 k-incracking at midspan
M M M
M M
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Serviceabil i ty Load Deflections - Example
Assume beam is fully cracked under full DL + LL,therefore I= Icr(do not calculate Iefor now).
Icr
for supports
2 2
s
2
2 2s
7.20
2 4 0.20 in 3 0.6 in
3.40 in
20 in - 2.5 in = 17.5 in
3 0.6 in 1.80 in
2.5 in
n
A
d
A
d
S i bi l i t L d D fl ti E l
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Serviceabil i ty Load Deflections - Example
Class formula using doubly reinforced rectangular section
behavior.
s s s s2
2 2
2
2 2
2 1 2 2 1 20
2 6.2 1.80 in 2 7.2 3.4 in
12 in
2 6.2 1.80 in 2.5 in 2 7.2 3.4 in 17.5 in 012 in
n A nA n A d nA d y y
b b
y y
Ser iceabil i t Load Deflections E ample
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Serviceabil i ty Load Deflections - Example
Class formula using doubly reinforced rectangular section
behavior.
s s s s2
2
2 1 2 2 1 20
5.94 76.05 0
5.94 3406.25 in.
2
n A nA n A d nA d y y
b b
y y
y
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Serviceabil i ty Load Deflections - Example
Calculate moment of inertia.
2 23
s scr +
3
22
22
4
11
3
1 12 in 6.25 in3
6.2 1.80 in 2.5 in 6.25 in
7.2 3.4 in 17.5 in 6.25 in
4230 in
I by n A y d nA d y
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Serviceabil i ty Load Deflections - Example
Weighted Icr
cr cr1 cr2cr mid
4 4 4
4
0.7 0.15
0.7 4330 in 0.15 4230 in 4230 in
4300 in
I I I I
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Serviceabil i ty Load Deflections - Example
Instantaneous Dead and Live Load Deflection.
4 34
D
DL inst 4
LL inst DL inst
1.00 k/ft 35 ft 12 in/ft
384 384 4030 ksi 4300 in
0.390 inches
0.62 k/ft*
1.00 k/ft
0.242 inches
l
EI
D
D D
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Serviceabil i ty Load Deflections - Example
Long term Deflection at the midspan
Dead Load (Duration > 5 years)
2
s
w
3 0.31 in0.00221
2 2 12 in 17.5 in
A tt
b d
DLDL L.T. DL inst
2.01.80
1 50 1 50 0.002211.8 0.390 in 0.702 in
l
l
D D
Serviceabil i ty Load Deflections Example
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Serviceabil i ty Load Deflections - Example
Long term Deflection use the midspan information
Live Load (40 % sustained 6 months)
2
s
w
3 0.31 in0.00221
2 2 12 in 17.5 in
A tt
b d
LLLL L.T. LL inst
1.21.08
1 50 1 50 0.00221
1.08 0.242 in 0.40
0.105 in
l
l
D D
Serviceabil i ty Load Deflections Example
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Serviceabil i ty Load Deflections - Example
Total Deflection after Installation of Glass Partition Wall.
total DL inst LL inst DL L.T. LL L.T.
after attachment total DL inst
permissible
0.390 in 0.242 in 0.702 in 0.105 in
= 1.44 in
1.44 in 0.105 in 1.33 in
480
35 ft 12 in/ft0.875 in < 1.33 in NO GOO
480
l
D D D D D
D D D
D
D!
Serviceabil i ty Load Deflections Example
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Serviceabil i ty Load Deflections - Example
Check whether modifying Icr
to Iewill give an acceptable
deflection:
3 3
cr cr g cre midspan
a a
3
4
34
4
1
618 k-in16900 in
1380 k-in
618 k-in1 4330 in
1380 k-in
5460 in
M MI I I
M M
Serviceabil i ty Load Deflections Example
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Serviceabil i ty Load Deflections - Example
Check whether modifying Icrto Iewill give an acceptable
deflection:
3
4
e support
3
4
4
1610 k-in16900 in
2000 k-in
1610 k-in1 4230 in
2000 k-in
10800 in
I
e1 e2e avg e mid
4 4 4
4
0.7 0.15
0.7 5460 in 0.15 10800 in 10800 in
7060 in
I I I I
Serviceabil i ty Load Deflections Example
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Serviceabil i ty Load Deflections - Example
Floor Beam meets the ACI Code Maximum permissible
Deflection Criteria. Adjust deflections:
4
DL inst 4
4
LL inst 4
4
DL L.T. 4
4
LL L.T. 4
4300 in0.390 in 0.238 in
7060 in
4300 in0.242 in 0.147 in7060 in
4300 in0.702 in 0.428 in
7060 in
4300 in0.105 in 0.064 in
7060 in
D
D
D
D
S i bi l i t L d D fl ti E l
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Serviceabil i ty Load Deflections - Example
Adjust deflections:
total DL inst LL inst DL L.T. LL L.T.
after attachment total DL inst
permissible
0.238 in. 0.147 in. 0.428 in. 0.064 in.
0.877 in.
0.877 in. 0.105 in.
0.772 in. 0.875 in. OKAY!
D D D D D
D D D
D
S i bil i t L d D fl ti E l
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Serviceabil i ty Load Deflections - Example
Part II: Check crack width @ midspan
3s cz f d A
c s
2
e s2
2e
2.5 in
A 2 2 2.5 in 12 in 60 in
60 in15 in
# bars 4
d d
d b
AA
S i bi l i t L d D fl ti E l
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Serviceabil i ty Load Deflections - Example
Assume
s y0.6 0.6 60 ksi 36 ksi ACI 10.6.4f f
2336 ksi 2.5 in 15 in
120 k/in < 175 k/in OK
z
For interior exposure, the crack width @ midspan
is acceptable.
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http://civilengineeringreview.com/book/theory-
structures/beam-deflection-method-superposition
http://911review.org/WTC/concrete-core.html
http://iisee.kenken.go.jp/quakes/kocaeli/index.htm